Is Haar measure reflection invariant?

Question

Let $G$ be a locally compact group and $\mu$ be a Haar measure on $G.$ Is $\mu$ necessarily reflection invariant i.e. can we always say that $\mu (E) = \mu (E^{-1})\ $?

where $E \subseteq G$ is a measurable set and $E^{-1} = \left \{x \in G\ |\ x^{-1} \in E \right \}.$

For Lebesgue measure I know that the result holds. But for Haar measure defined on an arbitrary locally compact group I don't know the result.

Answer 1

Note that, if $\chi_E$ is the characteristic function of $E$, then$$\mu(E)=\int_G\chi_E\,\mathrm d\mu\quad\text{and}\quad\mu\left(E^{-1}\right)=\int_G\chi_E\circ\iota\,\mathrm d\mu,$$with $\iota(x)=x^{-1}$. But, in general, we do not have$$\int_Gf(x)\,\mathrm d\mu(x)=\int_Gf(x^{-1})\,\mathrm d\mu(x);$$this happens only for the unimodular groups.

If you want a concrete example, take$$G=\left\{\begin{bmatrix}a&b\\0&1\end{bmatrix}\,\middle|\,a>0\wedge b\in\Bbb R\right\}.$$A left Haar measure for this group is$$\mu(E)=\iint_E\frac1{x^2}\chi_E\,\mathrm dx\,\mathrm dy.$$Now, take $E=[1,2]\times[0,1].$ Then $\mu(E)=\frac12$, but $\mu\left(E^{-1}\right)=\log(2)$.

Answer 2

Your assumption means that a left Haar measure is also a right Haar measure, so your assumption happens if and only if the locally compact group $G$ is unimodular.

So, for a concrete counterexample, consider any non-unimodular group such as the $ax+b$-group.